Every possible reaction in chemistry is to attain stability. In physics, the alignment of an electric dipole in an external electric field and in all other physical systems (at least those I study in high school) attains stability at the lowest energy state. But, why is it so?

6 Answers
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To answer your question, you should first understand when is a system most stable.

Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero.

We know that $$F = - \frac{dU}{dx}$$
Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$

Secondly, it should be able to maintain that equilibrium condition by itself. This can be tested by displacing the system by a small distance $dx$. If the force on the system then becomes opposite to direction of $dx$, we can say that the system has a tendency to restore back to its original equilibrium position.
An example of this would be a ball kept at the bottom of a spherical valley. Displace the ball a little towards the right, and the net force on it acts towards left, bringing it back to its original position. You will realise that I just described a stable equilibrium condition. What this proves is that it is the stable equilibrium condition in which the system is most stable.

From the above description we have that the small displacement $dx$ and net extra force $dF$ should be in opposite directions.

But, why is it that the lowest potential state corresponds to the state of stable equilibrium? Can you give an answer without using mathematical results for getting better intuition on the answer provided.
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Rajath Krishna RNov 21 '13 at 16:05

Minimizing Gibbs free energy under constant temperature and pressure is a standard result in thermodynamics textbook...
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user26143Nov 21 '13 at 16:49

I'll try to explain with the help of an classical example. Take the situations in the picture above. What you're interested in are the first to cases. The unstable state of equilibrium is such a state that when you slightly displace the ball, it departs from the original position. Being at the top of the hill, it has an excess of potential energy (may it be gravitational, electric, etc.) that can be converted into work. However, in stable equilibrium, displacing the ball it will always return to its original position. It was from the start in a state with minimum energy.

Well, chemical reactions almost always require heat (energy) to take a place, and almost always release heat upon reaction, so by that logic state when elements is unable to keep reacting is a state with insufficient energy or, in other words, lowest energy state (or we probably should say "lower energy state" then one that required for reactions)

A system that is in thermal contact with it's environment will tend towards both a lower energy state and a higher entropy state. Basically, the energy of the system + environment is fixed, but energy will flow between the two until they are in a state of maximum entropy.

It might be more informative to ask why systems tend towards increased entropy. What happens is that all the states that the system+environment can occupy with fixed total energy have equal probability of being occupied. This is called the fundamental postulate of statistical mechanics. Now there are many such states for which the system has some particular energy $E$. The value of $E$ that corresponds to the greatest number of states is therefore most likely.

if you look around using a constrained google search for filetype:pdf you can find it free.

That being said the lowest potential state is likely a lattice at absolute zero with a perfectly isomorphic electromagnetic state. Technically anything less than all matter reaching that state is subject to being just a local equilibria point awaiting a lower (more broadly contextualized) state i.e. ultimate heat death.